For $n\in\mathbb N$, $n!$ could, theoretically, be expanded into a polynomial of degree $n$ as $$\underbrace{n(n-1)(n-2)(n-3)\cdots \left(n-(n-2)\right) (n-(n-1))}_{n \ \text{factors}} =\sum_{k=0}^n a_k n^k $$ How can I determine the coefficients $a_k$?
For the $n^n$ term, there is only one choice, as every factor must contribute an $n$. So $a_n$ should be $1$.
For the $n^{n-1}$ term, we need $n-1$ factors to contribute an $n$, and the remaining factor multiples it with a constant term. So, $$a_{n-1} = -\sum_{i=0}^n i$$and so on. But I’m not sure if what I’m doing really makes sense. Does such a polynomial represention of $n!$ really exist?
You are describing the functions $(x)_n = x(x-1)\cdots(x-n+1)$ , which are known as the Falling Factorials. Expressed as polynomials, these have the Stirling Numbers of the First Kind as their coefficients.
To be specific, we have $(x)_n = \sum_{k=0}^n s(n,k) x^n$ where $s(n,k)$ denotes the $(n,k)$th Stirling Number of the First Kind. From the definition of $(x)_n$ we can, with a little tinkering, find that $s(n+1,k) = s(n,k+1) - n s(n,k)$ (see my comments above) which allows the coefficients to be recursively calculated as desired.